Question Video: Determining the Dot Product between Given Vectors | Nagwa Question Video: Determining the Dot Product between Given Vectors | Nagwa

# Question Video: Determining the Dot Product between Given Vectors Mathematics • Third Year of Secondary School

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Given that π¨ = β3π’ β 5π£ + π€, π© = β5π’ β 3π£ β 3π€, π = β2π’ β π£ + 4π€, and (π¨ + ππ©) is perpendicular to the vector π, determine π.

02:03

### Video Transcript

Given that vector π¨ equals negative three π’ minus five π£ plus π€, vector π© equals negative five π’ minus three π£ minus three π€, vector π equals negative two π’ minus π£ plus four π€, and π¨ plus π times π© is perpendicular to the vector π, determine π.

Given these three three-dimensional vectors π¨, π©, and π, we know that if we take vector π¨ and we add it to this multiple π of vector π©, then that result is perpendicular to vector π. Whenever two vectors are perpendicular to one another, it means something very specific. Because the two vectors π¨ plus π times π© and π are perpendicular, we can say that the dot product between them equals zero.

In this expression, weβre given all the values for our three vectors. But we donβt yet know the value of this multiple π. Thatβs what we want to solve for. And we can start to do that by substituting in the components of vectors π¨, π©, and π. When we do so, we get this big long expression. The part weβll focus on for now is where we have vector π¨ being added to this multiple π of vector π©.

As a first step, letβs multiply the components of π© by this factor π. Next, we can add together these two vectors, adding them component by component. For the π’-component, we have negative three minus five π; the π£-component is negative five minus three π; and then for the π€-component positive one minus three π.

Weβre now ready to go ahead and compute this dot product. We do this by multiplying together the π’-components, then adding that to the multiple of the π£-components, and adding that to the multiple of the π€-components. That gives us this expression. And working on the left-hand side, we can combine together six, five, and four β thatβs 15 β and 10π, three π, and negative 12π β thatβs simply π. And as weβve seen, this is equal to zero. That tells us that π equals negative 15.

So, given these vectors π¨, π©, and π, for π¨ plus π times π© to be perpendicular to π, π equals negative 15.

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